A
Possible Scalar Term Describing Energy Density in the Gravitational Field

The gravitational
field of a point mass and the electric field of a point charge are structurally
similar. Each may be represented by a vector field in which the vectors
are directed along radial lines emanating from the point, and the vector
magnitudes decrease as the inverse square of the [radial] distance
from the point. The electric field, E,
when multiplied by the magnitude of a test charge q in the field, gives
the force f
exerted locally on the test charge

f
= qE

1.

The gravitational
field, g,
when multiplied by the magnitude of a test mass m in the field, gives
the force f
exerted locally on the test mass

f
= mg

2.

In each
case, the force has the same vector sense as the field. The electric field
may point radially toward or away from the source charge, depending on
the sign of the source charge. The gravitational field points toward the
source mass in all known cases. The electric field has a scalar energy
density field (or,following some older texts, a pressure field) associated
with it. When the field vector at a point is E,
then the energy density at the same point is

uE
= ½(0E.E
) =
½0E2

3.

where
015
is the permittivity of free space. Energy density is a measure of the
energy stored in the field per unit volume of space. Its unit of measure
is j/m3 (or nt/m2, if it is thought of as a pressure).
While eq. 3 represents the energy density for the electric
field, and a similar expression represents the energy density for the
magnetic field, no such energy density term has ever been defined for
the gravitational field. But one suspects that it could be, and possibly
even should be.

Let us
use the similarity between the gravitational and electric fields to construct
a gravitational energy density term.

We begin
by noting how the permittivity of free space enters into the expression
for E:

E
= kQ/r2

4.

where
k = 1/(4p0)

4a.

is a
universal constant, Q is the source charge, and r is radial distance from
the source.

The gravitational
field is given by the expression

g
= GM/r2

5.

where
G is a universal constant, M is the source mass, and r is radial distance
from the source.

The electrical
field energy density may be written in terms of k as

uE
= ½ (1/4k)
E2

=
E2/(8k)

6.

By analogy,
a candidate gravitational energy density term may now be constructed and
written as

uG
= g2/(8G)

7.

Near
the surface of the earth

g =
9.807 m/sec2.

Also
G = 6.672 X 10-11 (nt m2)/kg2

so that
uG = 5.736 X 1010 j/m3.

Using
eq. 7, it might be possible to construct a classical
argument for the rotation of perihelion of a planet around its central
body [sun]. Recall that the rotation of Mercury's perihelion (43
arc sec per century) was successfully dealt with by general relativity.
Historically, attempts to modify Newton's law of gravitation to account
for the observed motion of Mercury have proved unsatisfactory. So did
the introduction of another [hypothetical] planet called Vulcan,
with an orbit inside that of Mercury. It is reasonable, then, that the
present argument will not be an exact solution, but a rough approximation.
Its usefulness is as a conceptual tool for students, familiar with classical
dynamics, who are just being introduced to Einstein's concepts. It will
serve the function of a bridge between classical orbit theory and general
relativity. Einstein's original paper on the deflection of starlight passing
the sun might be said to serve a similar function. Einstein's calculation
in that paper, is still reproduced as a problem in texts on special relativity.
Throughout this calculation, we will use Newton's law of gravitation without
modification, and incorporate the energy density term described above
to render a qualitative description of the rotation of perihelion.

Another
scalar field may be obtained from the energy density field by using the
mass-energy equivalence from special relativity; i.e., a scalar mass density
field of the form

uG/c2
= GM2/(8r4c2)

9.

We next
assume that the mass due to the term uG/c2, integrated
over a suitable volume of space, behaves, gravitationally, like ordinary
matter. For an extended body like the sun (rather than an ideal point
mass), the ramifications of this assumption need to be explored

i. In the
interior of the extended body, since the interior field contributes
to the overall mass, and, therefore, to its gravitational field, and

ii.
In the "matter free" space surrounding the extended body. Planets and
other objects moving in this region will experience the combined gravitational
effects of the mass called out in i. above, and, also, the mass contribution
of the external field.

The sun
may be considered a sphere of radius r0, and constant (average)
density .
The classical solar mass is then

M0
= 4r03/3

To find
the additional mass contribution due to the interior field, we must first
rewrite eq. 9 for the sun's interior. We set

M =
4r3/3

which
is a function of the radial distance r. We next substitute this result
into eq. 9 to obtain

uG/c2
= 2Gr22/(9c2)

10.

The corresponding
mass contribution Mf is then the volume integral of eq.
10 throughout the sun's interior. The integrand is

dMf
= [2Gr22/(9c2)]
. 4r2
dr

and the
limits of integration are from r = 0 to r = r0. Thus

Mf
= 82G2r05/(45c2)

11.

The total
mass of the sun must now be

M
= M0 + Mf

12.

For the
sun, we know that

M0
= 1.99 X 1030 kg
= 1,410 kg/m3 (avg.)
r0 = 6.912 X 108 m

We calculate
Mf = 4.08 X 1023 kg

As should
be expected, Mf is very small compared to M0. In
fact,

Mf
= (2.05 X 10-7) . M0

and is,
therefore, only slightly greater than one ten-millionth of the classical
solar mass.

Outside
the sun, the local field experienced by an orbiting planet, asteroid,
etc., has contributions from

i. The
mass M = M0 + Mf (a constant in this region),
and

ii.
The mass due to the external field, which will vary if the orbit is
not strictly circular.

Let a
planet be located a distance r from the center of the sun. Associate with
this distance a sphere, concentric with the sun, on whose surface the
planet is always to be found. As r increases, the sphere expands. As r
decreases, the sphere contracts. The motion of the planet is determined
by the total mass inside the sphere. This total mass is made up of the
mass contained in the sun plus the mass due to the external field contained
inside the sphere. For the external field

dMf'
= [GM2/(8r4c2)]
. 4r2
dr

The volume
integral is to be taken throughout the region between the surface of the
sun and the sphere of radius r. Thus

Mf'
= GM2(1/r0 - 1/r)/(2c2)

13.

It is
the total mass M + Mf' (= M0 + Mf + Mf')
that attracts the planet, and influences its motion around the sun. The
gravitational field g acting on the planet is

g
= G(M + Mf')/r2

14.

=
G(M + GM2(1/r0 - 1/r)/(2c2))/r2

14a.

This
field represents the acceleration of the planet in its orbit around the
sun. The classical contribution of the sun is represented by

GM/r2

15.

and results
in a stationary, elliptical orbit, as expected. The additional contribution
of the sun, due to its external gravitational field, is represented by

G2M2(1/r0
- 1/r)/(2c2r2)

16.

This
field varies with radial distance r, the more so the smaller the value
of r. Therefore, any planet, near the sun and in a noncircular orbit,
will experience a sideways perturbation from a strict classical orbit.
If, without this perturbation, the planet would follow a classical ellipse,
the perturbation must be such as to divert it sunwards from the ellipse
as it travels from perihelion to aphelion. The reverse must be true for
the other half of the orbit, with the result that the line of apsides
must turn [slowly] with the same directional sense as the orbital
motion of the planet (i.e., clockwise or counterclockwise, viewed from
a suitable vantage point)16.

It is
interesting to calculate the approximate magnitude of the mass contribution
in eq. 13 for the planet Mercury. Accordingly, we need
to know that for Mercury,

r =
5.75 X 1010 m

Also,
for the sun, M = 1.99 X 1030 kg

r0
= 6.91 X 108 m

And G
= 6.67 X 10-11 nt m2/kg2.

Then,

Mf'
= 2.10 X 1024 kg

17.

It is
interesting that the mass calculated in eq. 17 is roughly
equivalent to the mass of the earth. It was commented earlier that when
Mercury's rotation of perihelion was first observed, astronomers attempted
to account for it by postulating another planet called Vulcan, whose orbit
was inside that of Mercury and whose gravitational influence perturbed
Mercury's orbit. These observations were made late in the nineteenth century,
before the advent of special relativity. Vulcan was never found, as we
now know, and the perturbation in Mercury's orbit was finally accounted
for satisfactorily by Einstein.

150
= 8.8542 x 10-12 F/m.

16
It may be shown that motion, under a law of central attraction of the
general form

k/r2+k1+r318.

results
in the type of motion we have been describing here; i.e., that the perihelion
of a planet, moving under such a law of attraction, will rotate at a rate
proportional to the constant, k1 (Levi-Civita, pg. 396). The
law of attraction stated in eq. 14a, above, only approximates
eq. 18. We may rewrite eq. 14a.
as

The term
k1 is not a constant, as eq. 18 requires,
but is a linear function of the distance, r. Thus, our use of the energy
density term provides, at best, only a crude approximation to the perihelion
problem. None the less, it still seems reasonable that the approach described
above should have instructional value, providing a conceptual bridge between
the worlds of classical physics/special relativity on the one hand, and
general relativity on the other.

Line
of apsides: the line connecting the ascending and descending nodes of
the orbit.